© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the forces acting on the body
The body of mass m = 1\, \text{kg} rests on a horizontal surface. Let the applied force be \vec{F} at an angle \theta from the horizontal. The forces acting on the body are:
Weight of the body: mg (vertically downward)
Normal reaction from the floor: N (vertically upward)
Frictional force: f (along the floor, opposing motion)
Applied force: \vec{F} (making an angle \theta from the horizontal)
Step 2: Express the maximum static friction
The coefficient of static friction is given by \mu_s = \frac{1}{\sqrt{3}} . The maximum possible static friction is
f_{\text{max}} = \mu_s \, N.
Step 3: Determine the normal reaction
If \vec{F} has a vertical component F \sin \theta upward, it reduces the normal force. Hence,
N = mg - F \sin \theta.
Step 4: Write the condition for impending motion
For the block to just start moving, the horizontal component of the applied force F \cos \theta must exceed or equal the maximum friction:
F \cos \theta = f_{\text{max}} = \mu_s \bigl(mg - F \sin \theta \bigr).
Step 5: Rewrite and simplify the equation
Substitute \mu_s = \frac{1}{\sqrt{3}} into the above equation:
F \cos \theta = \frac{1}{\sqrt{3}} \bigl(mg - F \sin \theta \bigr).
Rearrange to isolate F :
F \cos \theta + \frac{F}{\sqrt{3}} \sin \theta = \frac{mg}{\sqrt{3}}.
Factor out F :
F \Bigl(\cos \theta + \frac{1}{\sqrt{3}} \sin \theta \Bigr) = \frac{mg}{\sqrt{3}}.
Hence,
F = \frac{\frac{mg}{\sqrt{3}}}{\cos \theta + \frac{1}{\sqrt{3}} \sin \theta}.
Step 6: Optimize to find the minimum F
To find the minimum value of F , we need to maximize the denominator
\cos \theta + \frac{1}{\sqrt{3}} \sin \theta . The maximum value of
a \cos \theta + b \sin \theta is \sqrt{a^2 + b^2} . Here, a = 1 , b = \tfrac{1}{\sqrt{3}} , so:
\sqrt{a^2 + b^2} = \sqrt{1^2 + \left(\frac{1}{\sqrt{3}}\right)^2}
= \sqrt{1 + \frac{1}{3}}
= \sqrt{\frac{4}{3}}
= \frac{2}{\sqrt{3}}.
Thus, the maximum of \cos \theta + \frac{1}{\sqrt{3}} \sin \theta is \frac{2}{\sqrt{3}} . Substituting this back,
F_\text{min} = \frac{\frac{mg}{\sqrt{3}}}{\frac{2}{\sqrt{3}}}
= \frac{mg}{2}.
Step 7: Substitute numerical values
Given m = 1\, \text{kg} and g = 10\, \text{m/s}^2 , we get
F_\text{min} = \frac{(1\, \text{kg}) \times (10\, \text{m/s}^2)}{2} = 5\, \text{N}.
Final Answer
The minimum force required is approximately 5\, \text{N}.
